
A man drives his car on the highway where the speed limit is \[60\text{ kmph}\] . He has to cover a distance of \[240\text{ km}\] at a uniform speed on the road. If he increases his speed by \[60\text{ kmph}\] , he can reach the destination \[1\text{ hour}\] earlier. What is his original speed?
A. \[20\text{ kmph}\]
B. \[5\text{ kmph}\]
C. \[\text{60 kmph}\]
D. \[30\text{ kmph}\]
Answer
585.9k+ views
Hint: We need to find the original speed with which the man drives. Given distance to cover \[240\text{ km}\] . Using \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\] , we will get an equation $t=\dfrac{240}{s}...(i)$ . Given that if speed is increased by \[60\text{ kmph}\] , then the man can reach the destination \[1\text{ hour}\] earlier. Hence, $s+20=\dfrac{d}{t-1}$ . Using equation $(i)$ in this equation, the speed can be determined.
Complete step by step answer:
We need to find the original speed with which the man drives.
Given that the highway the speed limit \[=60\text{ kmph}\]
Distance to cover is denoted as $d$ and is given by
\[d=240\text{ km}\]
Let $s$ be the speed and $t$ be the time to cover.
We know that \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\]
$\Rightarrow s=\dfrac{d}{t}$
Given that \[d=240\text{ km}\] . Therefore,
$s=\dfrac{240}{t}$
From this, we get
$t=\dfrac{240}{s}...(i)$
Given that if speed is increased by \[60\text{ kmph}\] , then the man can reach the destination \[1\text{ hour}\] earlier.
$\Rightarrow s+20=\dfrac{d}{t-1}$
Substituting \[d=240\text{ km}\] in the above equation, we get
$s+20=\dfrac{240}{t-1}$
Rearranging the terms, we get
$t-1=\dfrac{240}{s+20}$
$\Rightarrow t=\dfrac{240}{s+20}+1$
Substituting equation $(i)$ in the above equation, we get
\[\dfrac{240}{s}=\dfrac{240}{s+20}+1\]
Rearranging the terms, we get
\[\Rightarrow 240(s+20)=s(240+s+20)\]
Simplifying further, we get
\[\Rightarrow 240s+4800=240s+{{s}^{2}}+20s\]
\[\Rightarrow {{s}^{2}}+20s-4800=0\]
Now let us solve this.
\[(s+80)(s-60)=0\]
From this,
\[(s+80)=0,(s-60)=0\]
Hence the values are
\[s=-80,s=60\]
Speed cannot be negative. Hence the answer is $s=60\text{ kmph }$ .
Hence, the correct option is C.
Note:
The equation \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\] will be used in these types of questions. Be careful with the units given. At the end, speed is taken as a positive value hence the negative value of speed doesn’t exist. We can observe that the speed traveled is within the given highway speed limit.
Complete step by step answer:
We need to find the original speed with which the man drives.
Given that the highway the speed limit \[=60\text{ kmph}\]
Distance to cover is denoted as $d$ and is given by
\[d=240\text{ km}\]
Let $s$ be the speed and $t$ be the time to cover.
We know that \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\]
$\Rightarrow s=\dfrac{d}{t}$
Given that \[d=240\text{ km}\] . Therefore,
$s=\dfrac{240}{t}$
From this, we get
$t=\dfrac{240}{s}...(i)$
Given that if speed is increased by \[60\text{ kmph}\] , then the man can reach the destination \[1\text{ hour}\] earlier.
$\Rightarrow s+20=\dfrac{d}{t-1}$
Substituting \[d=240\text{ km}\] in the above equation, we get
$s+20=\dfrac{240}{t-1}$
Rearranging the terms, we get
$t-1=\dfrac{240}{s+20}$
$\Rightarrow t=\dfrac{240}{s+20}+1$
Substituting equation $(i)$ in the above equation, we get
\[\dfrac{240}{s}=\dfrac{240}{s+20}+1\]
Rearranging the terms, we get
\[\Rightarrow 240(s+20)=s(240+s+20)\]
Simplifying further, we get
\[\Rightarrow 240s+4800=240s+{{s}^{2}}+20s\]
\[\Rightarrow {{s}^{2}}+20s-4800=0\]
Now let us solve this.
\[(s+80)(s-60)=0\]
From this,
\[(s+80)=0,(s-60)=0\]
Hence the values are
\[s=-80,s=60\]
Speed cannot be negative. Hence the answer is $s=60\text{ kmph }$ .
Hence, the correct option is C.
Note:
The equation \[\text{speed}=\dfrac{\text{distance}}{\text{time}}\] will be used in these types of questions. Be careful with the units given. At the end, speed is taken as a positive value hence the negative value of speed doesn’t exist. We can observe that the speed traveled is within the given highway speed limit.
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